Quantitative Aptitude Questions Asked in IBPS Clerk Main 2018-19 | Check Here

Dear Aspirants,

Quantitative Aptitude Questions Asked in IBPS Clerk Main 2018-19

IBPS Clerk Mains exam has concluded and many Aspirants had appeared for the same. Here are some Questions of Quantitative Aptitude based in the IBPS Clerk Mains exam 2018-19. Practice these Questions as these will help you get familiar with the questions for the upcoming Exams. 

Q1. A is twice as efficient as B. Both can complete a work together in  days. Quantity 1: Time taken by B to complete the work alone. 
Quantity 2: If C is 50% more efficient than A, then time taken by C to complete the work alone.

Quantity I > Quantity II

Quantity I ≥ Quantity II

Quantity II > Quantity I

Quantity II ≥ Quantity I

Quantity I = Quantity II or Relation cannot be established

Solution:

Q2. One of the roots of 2x² + bx - 5 = 0 is 1. 
Quantity 1: Value of the other root. 
Quantity 2: 2.5

Quantity I > Quantity II

Quantity I ≥ Quantity II

Quantity II > Quantity I

Quantity II ≥ Quantity I

Quantity I = Quantity II or Relation cannot be established

Solution:

Q3. Two dices are rolled simultaneously. 
Quantity 1: Probability that the sum of the numbers that appeared is a multiple of 5. 
Quantity 2: 1/6

Quantity I > Quantity II

Quantity I ≥ Quantity II

Quantity II > Quantity I

Quantity II ≥ Quantity I

Quantity I = Quantity II or Relation cannot be established

Solution:

From I,
Fovorable cases
= (1,4),(2,3),(3,2),(4,1),(5,5),(4,6),(6,4)
= 7
∴ Required prob. =7/36
From II, Quantity II =1/6
Quantity I > Quantity II

Q4. In a match of 50 overs, team A’s average runs for first thirty overs was 4.5 runs/over while for the remaining 20 overs the average was 5.5 runs/over. Team B chased the target and lost by 10 runs. Find the average runs per over scored by team B. (team B played all the 50 overs).

4.4

5.2

4.7

5.6

3.8

Solution:

Total runs scored by team B = 30 × 4.5 + 20 × 5.5 - 10
= 235
∴ Required answer = 235/50
= 4.7

Q5. In a bag which contains 40 balls there are 18 red balls and some green and blue balls. If two balls are picked from the bag without replacement, then the probability of the first ball being red and second being green is 3/26. Find the number of blue balls in the bag.

16

12

10

14

8

Solution:

Directions (6-10): In each of these questions, two equations numbered I and II are given. You have to solve both the equation and give answer 

Q6. I. 2x² +11x+15=0 
II. 4y²+13y+9=0

if x < y

if x ≤ y

if x > y

if x ≥ y

if x = y or the relationship cannot be established

Solution:

Q7. I. x²-36x+324=0 
II. y²-35y+216=0

if x < y

if x ≤ y

if x > y

if x ≥ y

if x = y or the relationship cannot be established

Solution:

Q8.

if x < y

if x ≤ y

if x > y

if x ≥ y

if x = y or the relationship cannot be established

Solution:

Q9. I. 2x² + 17x + 35 = 0 
II. 3y² + 17y + 24 = 0

if x < y

if x ≤ y

if x > y

if x ≥ y

if x = y or the relationship cannot be established

Solution:

I. 2x² + 17x + 35 = 0
2x² + 10x+ 7x + 35 = 0
2x (x+ 5) +7 (x+ 5) = 0
(2x+ 7) (x+ 5) =0
x = (-7)/2, -5
II. 3y² + 17y + 24 = 0
3y² + 9y + 8y + 24 = 0
3y (y + 3) + 8 (y+ 3) = 0
(y+ 3) (3y + 8) = 0
y= -3, -8/3
y > x

Q10. I. x² + 72 = 108 
II. y³ + 581 = 365

if x < y

if x ≤ y

if x > y

if x ≥ y

if x = y or the relationship cannot be established

Solution:

I. x² + 72 = 108
x² =108 – 72 = 36
x = ±6
II. y³ + 581 = 365
y³ = –216
y = –6
x ≥ y

               

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